Question

For an isentropic flow of a perfect gas, an increase in Mach number results in:

• A. Increase in temperature
• B. Decrease in temperature
• C. Constant temperature
• D. Infinite temperature
• E. Zero temperature

Hint:

Static temperature is a measure of random molecular motion. When flow speeds up (higher Mach), more energy is directed forward, leaving less for random motion (lower temperature).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In gas dynamics, isentropic flow refers to a flow that is both adiabatic (no heat transfer) and reversible. In such a system, the total energy (stagnation temperature) remains constant.
Step 2: Key Formula or Approach:
The relationship between static temperature (\( T \)), stagnation temperature (\( T_0 \)), and Mach number (\( M \)) for a calorically perfect gas is:
\[ \frac{T_0}{T} = 1 + \frac{\gamma - 1}{2} M^2 \]
Step 3: Detailed Explanation:
In an isentropic flow, \( T_0 \) is constant along the streamline.
Rearranging the formula for static temperature \( T \):
\[ T = \frac{T_0}{1 + \frac{\gamma - 1}{2} M^2} \]
As the Mach number (\( M \)) increases (meaning the gas is accelerating), the denominator \( (1 + \frac{\gamma - 1}{2} M^2) \) increases because \( \gamma \) (ratio of specific heats, \( \approx 1.4 \) for air) is greater than 1.
Since the denominator increases and the numerator is constant, the resulting value of \( T \) must decrease.
Physically, the thermal energy of the gas (temperature) is being converted into kinetic energy (velocity/Mach number).
Step 4: Final Answer:
An increase in Mach number leads to a decrease in static temperature.